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Constructive and Destructive Interference

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Two identical loudspeakers are located at points A & B, 2.00 m apart. The loudspeakers are driven by the same amplifier and produce sound waves with a frequency of 784 Hz. Take the speed of sound in the air to be 344m/s. A small microphone is moved out from point B along a line perpendicular to the line connecting A and B.

    (a) At what distances from B will there be constructive interference?
    (b) At what distances from B will there be destructive interference?
    (c) If the frequency is made low enough, there will be no positions along the line BC at which destructive interference occurs. How low must the frequency be for this to be the case?


    2. Relevant equations

    v=fλ
    f=nv/2L

    3. The attempt at a solution

    I'm not sure how to approach this problem. All I know is that constructive interference occurs when the distances traveled by the two waves differ by a whole number of wavelengths, whereas destructive interference occurs when the distances traveled differ by a half-integer number.

    Any direction would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Dec 6, 2011 #2
    First step: find the wavelength λ.
     
  4. Dec 6, 2011 #3
    Ok, that's what I was thinking. And I'd want to use λ=v/f, but I don't have a value for v -- other than the speed of sound through the air but that isn't necessarily v for the sound emitted by the speakers...
     
  5. Dec 6, 2011 #4
    The speakers are emitting sound. This sound is travelling through the air. The interference is interference between sound waves.
    So it is the speed of sound that one requires.
     
  6. Dec 6, 2011 #5
    Ok. So, it is appropriate to use 344m/s when talking about speed of sound through the air, but if we're talking about the speed of wave propagation of say a guitar string, it will not necessarily be that same value.

    So I calculated .439m for the wavelength...and I'm really not sure where to go from here.
     
  7. Dec 6, 2011 #6
    Do you know the condition for constructive interference in terms of path difference and wavelength.
     
  8. Dec 6, 2011 #7
    constructive interference: path difference is nλ for integer n and
    destructive interference: path difference is (n + 1/2)λ

    ... as you said.
     
  9. Dec 6, 2011 #8
    I think so. For constructive interference the distances traveled by the two waves differ by a whole number of wavelengths. And for destructive interference, they differ by half-integer numbers.

    So the wavelength I solved for in this case, is the same wavelength for both speakers A and B? And if that's the case, then for part (a), is it simply 0, 1*.439, 2*.439, 3*.439...etc, or am I missing a step?

    Then I would think for destructive interference, it would occur at distance .439/2, [(3)(.429)]/2, [(4)(.439)]/2...etc.
     
  10. Dec 6, 2011 #9
    Can you write down an expression for the path difference from A and B?
     
  11. Dec 6, 2011 #10
    If the distance from B to the first max is x, what is the distance from A to this first max?
    remember B is on a line perpendicular to the line AB
     
  12. Dec 6, 2011 #11
    Ok, so the distance from B is x. The distance from A would be [itex]\sqrt{x^{2}+2^{2}}[/itex] = the distance from A to x?
     
  13. Dec 6, 2011 #12
    A to x must be 1 wavelength more than x for the first max so
    the distance from A to x = x + 0.439
     
  14. Dec 6, 2011 #13
    Why is that? I don't understand that
     
  15. Dec 6, 2011 #14
    You are almost there now.
    Waves from A meet waves from B to reinforce. The first time this happens is when the path difference is one wavelength so the distance from A to the max must be 1 wavelength greater than the distance from B.
    A, B and the max form a right angle triangle so pythagoras relates the distances.
    x^2 + 2^2 = (x+0.439)^2..... should give you x
     
  16. Dec 6, 2011 #15
    I got x = 4.34m
     
  17. Dec 6, 2011 #16
    I got 4.34m for x using the above method you mentioned, and I'm not entirely sure where to go from there...Thoroughly confused about the subject matter, you could say...
     
  18. Dec 6, 2011 #17

    Delphi51

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    Homework Helper

    You could do it again with 2λ, then 3λ, ...
    Or do it once with path difference nλ and evaluate x for n = 1, 2, 3, ...
    And again with (n+.5)λ.
     
  19. Dec 6, 2011 #18
    I'm sorry, I don't understand. My thought is to simply do (.439)2, (.439)3. I don't see how speaker A comes into play, since they both have the same wavelength.
     
  20. Dec 6, 2011 #19

    Delphi51

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    Homework Helper

    Oh, you want the path difference between A to x and B to x.
    That's sqrt(x² + 4) and x. Set that difference equal to nλ and solve for x. Evaluate for as many values of n as you need.
     
  21. Dec 6, 2011 #20
    Ok, so solving for x I got 4.34m. Then plugging that back into sqrt(x^2+4) I find the distance between A and x, which is 4.78. Finding the difference -- 4.78-4.34=.44m. Is that the value I use for the wavelength? I still don't seem to be getting the correct answers.
     
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